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Math Help - Statistics - Esitmate Mean,Median,Standard deviation using Freq. table.

  1. #1
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    Statistics - Esitmate Mean,Median,Standard deviation using Freq. table.

    hello!

    I am wondering if I could please get help with the below questions which are revision questions but I do not know how to do it.

    Additional Problems: Grouped Data
    1. Estimate the mean, median and the standard deviation of the following set of
    grouped data that summarise the marks (out of 100) obtained in a statistics
    exam for a sample of 150 computer science students.

    Score...............Frequency
    10- up to 20.........1
    20- up to 30.........6
    30- up to 40.........9
    40- up to 50.........31
    50- up to 60.........42
    60- up to 70.........32
    70- up to 80.........17
    80- up to 90.........10
    90- up to 100.........2

    for the mean do you calculate the fimi total and divide that by the total freq. ?
    for the median do you just divide the total freq. by 2 ? which gives us the m class

    We can estimate the median by using L + j/f c

    and for the standard deviation, I have no idea... ?

    2. The gross hourly wages of a sample of 100 workers randomly selected from a
    payroll list of a large company were tabulated as follows:

    Hourly wages($)...................No: ofworkers

    8- up to 10..................................11
    10- up to 12................................17
    12- up to 14................................32
    14- up to 16................................27
    16- up to 18................................13

    Estimate the mean, median and the standard deviation of hourly wages for this
    sample of workers

    would be the same formula's as the previous question.
    Last edited by Jon123; May 20th 2011 at 07:57 AM. Reason: Statistics
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  2. #2
    Senior Member Sambit's Avatar
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    For mean: \frac{sum\quad of\quad (class\quad mark*frequency)}{sum\quad of frequency} where
    class\quad mark=\frac{lower\quad bound+upper\quad bound}{2}

    For median: Check where the cumulative frequency attains half-frequency-value. The corresponding class (or the class-mark) is the median.

    For the first problem, mean is 57.2 and median-class is 50-60 and the standard deviation is 15.31318

    The excel-sheet of first problem is attached. Do the second problem in same way.New Microsoft Excel Worksheet (2).xls
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Sambit View Post
    For mean: \frac{sum\quad of\quad (class\quad mark*frequency)}{sum\quad of frequency} where
    class\quad mark=\frac{lower\quad bound+upper\quad bound}{2}

    For median: Check where the cumulative frequency attains half-frequency-value. The corresponding class (or the class-mark) is the median.

    For the first problem, mean is 57.2 and median-class is 50-60 and the standard deviation is 15.31318

    The excel-sheet of first problem is attached. Do the second problem in same way.New Microsoft Excel Worksheet (2).xls
    The class marks you have used are too low by 0.5, as the lower end point is not included and only integer marks (presumably) are permitted (if half marks that would be 0.25 ratehr than 0.5)

    (possibly too high by 0.5 if it is the upper point that belongs to the next class rather than the lower to the previous, I think the notation may make this the more likely interpretation)

    CB
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  4. #4
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    Hi thanks for replying!

    With the standard deviation part, could you please provide your working out for you getting 15.31318

    Ive tried to get that answer but cant. when I times 'f' by 'fx' i get a total of 241,130
    and when i divide that by 150 i get 1607.53

    then 1607.53 - 57.2^2
    = sqrt of -1,664.30

    just isnt the same answer.... could you please help,,
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  5. #5
    Senior Member Sambit's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The class marks you have used are too low by 0.5, as the lower end point is not included and only integer marks (presumably) are permitted (if half marks that would be 0.25 ratehr than 0.5)

    (possibly too high by 0.5 if it is the upper point that belongs to the next class rather than the lower to the previous, I think the notation may make this the more likely interpretation)

    CB
    The classes in the problem are 10-20,20-30,30-40 -etc which means the lower boundary of one class is the upper boundary of the next class. So class marks are obtained simply by taking the mid-value of the lower and upper boundaries of each class.
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  6. #6
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    for the median I ended up getting 56.6

    please let me know if this is correct
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  7. #7
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Jon123 View Post
    Hi thanks for replying!

    With the standard deviation part, could you please provide your working out for you getting 15.31318

    Ive tried to get that answer but cant. when I times 'f' by 'fx' i get a total of 241,130
    and when i divide that by 150 i get 1607.53

    then 1607.53 - 57.2^2
    = sqrt of -1,664.30

    just isnt the same answer.... could you please help,,
    You can' do that since you need to get class-mark^2 * frequency that is x*x*f, not f*f*x.
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  8. #8
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Jon123 View Post
    for the median I ended up getting 56.6

    please let me know if this is correct
    Where are you getting that value from? Finding median means finding the class-mark of the class corresponding to which the cumulative frequency exceeds half-frequency (here half of 150 that is 75). So the median is class-mark of the class 50-60 that is 55.
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  9. #9
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    Hi Sambit,

    I am using the formula

    Median (approx)= L + j/f . c
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  10. #10
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Jon123 View Post
    We can estimate the median by using L + j/f c
    Sorry I forgot something. You probably mean this formula: median = l+\frac{\frac{n}{2}-F_l}{f_{median}} \into c.

    In that case median will be 56.67; so you are correct.
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